The minimum value of the function \(f(x) = 3 \sin x-4 \cos x, x \in [-4\pi, 4\pi]\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −5
An expression \(a\sin x + b\cos x\) can be written as a single sine, \(R\sin(x - \alpha)\), with \(R = \sqrt{a^2 + b^2}\). Here
\(\displaystyle R = \sqrt{3^2 + 4^2} = 5\)
Let \(\alpha\) be the acute angle with
\(\displaystyle \cos\alpha = \frac{3}{5},\quad \sin\alpha = \frac{4}{5}\)
Then
\(\displaystyle f(x) = 5\left(\frac{3}{5}\sin x - \frac{4}{5}\cos x\right)\)
\(= 5(\sin x\cos\alpha - \cos x\sin\alpha) = 5\sin(x - \alpha)\)
Since \(-1 \le \sin(x - \alpha) \le 1\),
\(-5 \le f(x) \le 5\)
The value −5 is reached when \(x - \alpha = -\frac{\pi}{2}\), that is at \(x = \alpha - \frac{\pi}{2}\), which lies in \([-4\pi, 4\pi]\). So the minimum value is −5.
Why the other options are incorrect
Option (1) 0 — Incorrect. f(x) = 0 when \(\tan x = \frac{4}{3}\), but f takes smaller values, for example f(0) = −4.
Option (2) −4 — Incorrect. This is f(0) = 3(0) − 4(1), which is not the smallest value.
Option (4) −7 — Incorrect. This would need \(\sin x = -1\) and \(\cos x = 1\) at the same x, which is impossible since \(\sin^2 x + \cos^2 x = 1\).